suppose A = {a_{1},...,a_{n}} is a basis for a vector space V.

suppose further that k ≠ 0 in the field F that V is a vector space over.

is B = {ka_{1},...,ka_{n}} also a basis? let's see:

suppose c_{1}(ka_{1}) +...+ c_{n}(ka_{n}) = 0.

then k(c_{1}a_{1} +...+ c_{n}a_{n}) = 0.

since k ≠ 0, c_{1}a_{1} +...+ c_{n}a_{n} = 0.

by the linear independence of the a_{j}, c_{1} =...= c_{n} = 0.

thus the ka_{j} are also linearly independent.

we know that the a_{j} span V. this means given ANY v in V we can write:

v = c_{1}a_{1} +...+ c_{n}a_{n} for some c_{j} in F.

thus v = (c_{1} /k)(ka_{1}) +....+ (c_{n}/k)(ka_{n}), so the ka_{j} span V as well.

now, suppose that each a_{j} = p_{j}/q_{j}, where the p's and q's are relatively prime integers for each j.

what about k = lcm(q_{1},..,q_{n})? this certainly gives a basis that is all integers.

however, i don't believe you can realize the span of this basis as Z-linear combinations, only as Q-linear combinations.

let me give an example:

suppose we have the matrix:

\(\displaystyle A = \begin{bmatrix}2&1&3\\4&-5&1\\6&-4&4 \end{bmatrix} \)

which has a null space with basis {(8,5,-7)}. now (1,5/8,-7/8) is also in this null space, but NO Z-multiple of (8,5,-7) will ever give us (1,5/8,-7/8).